(18) Out of 120 students in SSS 3, [tex]$\frac{5}{8}$[/tex] like football and [tex]$\frac{2}{5}$[/tex] like swimming. Every student like at least one of these sports.
(a) What fraction of the students in [tex]$SSS _3$[/tex] like both subjects?
(b) What fraction of the students like football but not swimming?
(c) What fraction of the students like swimming only?
Answer (1)
Calculate the number of students who like football: 8 5 × 120 = 75 .
Calculate the number of students who like swimming: 5 2 × 120 = 48 .
Use the inclusion-exclusion principle to find the number of students who like both: 120 = 75 + 48 − n ( F ∩ S ) , so n ( F ∩ S ) = 3 .
Determine the fractions: Both: 120 3 = 40 1 , Football only: 120 75 − 3 = 5 3 , Swimming only: 120 48 − 3 = 8 3 .
40 1 , 5 3 , 8 3
Explanation
Understand the problem We are given that there are 120 students in SSS 3. We also know that 8 5 of the students like football and 5 2 of the students like swimming. Every student likes at least one of these sports. We need to find the fraction of students who like both sports, the fraction who like football but not swimming, and the fraction who like swimming only.
Define sets and objectives Let F be the set of students who like football, and S be the set of students who like swimming. We are given the following information:
Total number of students = 120
Fraction of students who like football = 8 5
Fraction of students who like swimming = 5 2
Every student likes at least one of the sports. This means n ( F ∪ S ) = 120
We want to find:
The fraction of students who like both sports, i.e., 120 n ( F ∩ S )
The fraction of students who like football but not swimming, i.e., 120 n ( F ) − n ( F ∩ S )
The fraction of students who like swimming only, i.e., 120 n ( S ) − n ( F ∩ S )
Calculate number of students who like football and swimming First, let's find the number of students who like football and swimming:
n ( F ) = 8 5 × 120 = 75
n ( S ) = 5 2 × 120 = 48
Calculate number of students who like both sports Using the principle of inclusion-exclusion, we have:
n ( F ∪ S ) = n ( F ) + n ( S ) − n ( F ∩ S )
Since every student likes at least one sport, n ( F ∪ S ) = 120 . Plugging in the values, we get:
120 = 75 + 48 − n ( F ∩ S )
n ( F ∩ S ) = 75 + 48 − 120 = 3
Calculate fraction of students who like both sports (a) The fraction of students who like both sports is:
120 n ( F ∩ S ) = 120 3 = 40 1
Calculate fraction of students who like football but not swimming (b) The number of students who like football but not swimming is:
n ( F ) − n ( F ∩ S ) = 75 − 3 = 72
The fraction of students who like football but not swimming is:
120 72 = 60 36 = 10 6 = 5 3
Calculate fraction of students who like swimming only (c) The number of students who like swimming only is:
n ( S ) − n ( F ∩ S ) = 48 − 3 = 45
The fraction of students who like swimming only is:
120 45 = 24 9 = 8 3
State the final answer Therefore, the fraction of students who like both sports is 40 1 , the fraction of students who like football but not swimming is 5 3 , and the fraction of students who like swimming only is 8 3 .
Examples
Understanding sports preferences in a school can help in planning extracurricular activities. For example, if a school knows that a large fraction of students likes football but not swimming, they might invest more in football facilities. Conversely, if many students like both, the school could organize combined sports events to maximize participation and enjoyment.
